Download Problem Set 3 Answer Key, Spring 2003 1) The following

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Transcript
Problem Set 3 Answer Key, Spring 2003
1) The following questions refer to important general features of receptor tyrosine kinase
and MAPK pathways.
A) Describe the structural features of a receptor tyrosine kinase.
1) extracellular ligand binding domain
2) single hydrophobic transmembrane _-helix
3) cytosolic signaling domain with
a) kinase domain
b) tyrosine residue(s) in the (phosphorylation lip) near the catalytic site that
become phosphorylated upon ligand binding and receptor dimerization
allowing full kinase function
c) tyrosine residues elsewhere that are phosphorylated by the kinase domain
and promote binding of other proteins
B) What is the purpose of Ste5 in the yeast mating pathway?
Ste5 functions as a scaffold to assemble the multiple kinase components of the
pathway. This helps to ensure specificity since some of the kinases in the pathway
have other targets and can mediate alternative downstream responses when not
bound to Ste5. For example Ste11 is involved in both the mating and
osmoregulatory pathway and binding to a scaffold ensures only the desired signal
pathway is activated.
C) What are some differences between the switch protein GTPases G_ and Ras?
1) Ras requires a GAP to function as an effective GTPase whereas G_ does not
because it has a helical domain that serves this function.
2) Due to difference #1 G_ lasts in its GTP bound active state for a much shorter
time than Ras in general though a Ras-GAP can shorten the time for Ras.
3) G_ binds directly to its receptor whereas Ras is signaled through Grb2 and Sos.
4) Ras is directly anchored to the inner membrane.
5) Ras needs a GEF
2) You are studying the response of a fibroblast cell line to treatment with fibroblast
growth factor which you know binds to and acts through a receptor tyrosine kinase. The
cell line grows and divides when treated with growth factor, but do not in absence of
growth factor.
A) You express a mutant form of the fibroblast growth factor receptor lacking its entire
cytoplasmic domain to a level similar to the wild type receptor. You observe that the
cells grow much more slowly than normal when treated with growth factor. Explain.
The mutant receptor can bind ligand, but cannot signal. RTK’s dimerize when
binding ligand and are activated by transphosphorylation. Thus complexes with
one mutant and one wild type receptor will fail to signal. Only complexes of two
wild type receptors will signal, and these will be present at one quarter of the
normal levels so growth will be slow.
B) You inject anti-Ras antibodies that prevent Ras from binding Raf into a few cells.
What is the growth and division phenotype of these cells? Why?
The Ras antibodies bind to Ras and prevent its function. Thus it is unable to
activate the MAPK pathway, and the cells do not induce gene expression of growth
and division genes. The cells will not grow even in the presence of growth factor.
C) You express a constitutively active Ras that remains in its GTP bound form. You
observe that the cells divide and grow even in the absence of growth factor. Explain.
Ras signals to the downstream MAPK pathway whenever it is bound to GTP. Since
this mutant cannot hydrolyze GTP it continually signals and the cells divide despite
the lack of growth factor.
D) You treat the cells with RPR-130401, a chemical that inhibits farnesyl transferase,
and observe the effect on the cell growth and division. What do you see and why?
Ras has a farnesylation sequence which is farnesylated by farnesyl transferase. The
farnesyl group targets Ras to the membrane where it functions. Without a farnesyl
group Ras will be located throughout the cytoplasm and will not be efficiently
activated. Thus the cells will not grow even in the presence of growth factor.
E) You combine the activated Ras mutation from part C with the RPR-130401 treatment
from part D. What is the cellular growth phenotype?
Despite the constitutively active Ras without the farnesylation Ras cannot efficiently
signal to Raf and activate the MAPK pathway. Raf is activated primarily by being
brought to the membrane so if Ras is not at the membrane this will not occur. Thus
the cells will again not grow as in part D.
3) You are studying the Wg/Wnt pathway in a cell line, which responds to a Wnt peptide
treatment by fifteen-fold induction of the alkaline phosphatase gene.
You make several mutants of proteins in Wnt pathway components and overexpress them
(20fold more than wild type) in the cell line. You observe the relative level of alkaline
phosphatase mRNA before and after Wnt treatment. For the following scenarios describe
the ratio of alkaline phosphatase mRNA after to before Wnt treatment. Explain.
A) You overexpress a B-catenin mutant that has a serine to alanine mutation thatprevents
its phosphorylation by GSK3.
The mutant B-catenin cannot be inhibited by GSK3 so it will always be active and
induce the expression of the alkaline phosphatase gene. Thus the ratio of alkaline
phosphatase mRNA before and after Wnt treatment will be ~1.
B) You overexpress a GSK3 mutant that is unable to be bound by disheveled.
The mutant GSK3 cannot be inhibited by disheveled so it will always be active and
phosphorylated B-catenin. Thus B-catenin will never be active and the ratio of
mRNA will be ~1 since there will be little expression before or after Wnt treatment.
C) You overexpress a TCF mutant lacking its domain that interacts with B-catenin.
The mutant TCF cannot be bound by B-catenin so expression of alkaline
phosphatase can never be turned on and the ratio of mRNA before and after
treatment will be ~1.
D) You overexpress a B-catenin mutant that has a serine to glutamate mutation thus
mimicking its phospohorylated state.
The mutant B-catenin will be unable to enter the nucleus and act to induce gene
expression. However, it will not be bound by GSK3 since it is already
“phosphorylated”. Thus the wild type B-catenin in the cell can still respond to Wnt
treatment and the ratio of alkaline phosphatase mRNA after to before will remain
at ~15.
4) Cell junctions stabilize interactions and promote local communication between
adjacent cells.
A) Fill in the following table for each type of cell junction:
Cell junction
Cell-cell or cellExtracellular
ECM interaction?
interaction
Tight junction
Cell-cell
Occludin & claudin
from adjacent cells
Adherens
junction
Cell-cell
E-cadherins from
adjacent cells
Desmosomes
Cell-cell
desmoglein and
desmocollin from
Intracellular
interaction
Membrane occludin
– ZO-1/2/3 – actin
fibers
Membrane Eβcadherin – α/β
catenins – actin and
myosin belt
Plakoglobin plaques
– desmoplakin –
Function
Impermeable
seal b/w
basolateral
and apical
membranes
Rigid cell-cell
attachment
Strong cellcell
cadherin family)
Connexin hemichannels from
adjacent cells
keratin filaments
---
Membrane integrins
– plectin – keratin
filaments
Membrane integrin
– vinculin – actin
cytoskeleton
Gap junction
Cell-cell
Hemidesmosomes
Cell-ECM
Cellular integrins –
ECM laminin
Focal adhesion
Cell-ECM
Cellular integrin –
ECM fibronectin
attachment
Movement of
ions & small
molecules
b/w adjacent
cytosols
ECM
attachment
ECM
attachment
B) Specify how Ca2+ affects each of these cell junctions.
Tight junctions are disrupted in the absence of Ca2+ in the medium (unknown
mechanism).
Adherens junctions and desmosomes are disrupted in the absence of Ca2+ in the
medium since these structures contain members of the cadherin family and these
are Ca2+ - dependent.
The channels of gap junctions are closed in the presence of high intracellular Ca2+
concentrations.
As far as we know, hemi-desmosomes and focal adhesion are not affected by Ca2+.
5) You have a mutant mouse line that cannot fight infections and die shortly thereafter.
You suspect these mice have a problem in leukocyte extravasation which allows
leukocytes to move from the bloodstream into the infected tissues. Name two interactions
of cell adhesion molecules (CAMs) that could be defective in these mice. Indicate to
what class each CAM belongs to and where each CAM is normally found.
P-selectin (Selectin - endothelial cell membrane) – Sialyl Lewis-x antigen (Mucin leukocyte membrane)
αLβ2 integrin (Integrin - leukocyte membrane) – ICAM-1/2 (Ig-superfamily endothelial cell membrane)